Question

Find y' if 5y + 8x^(4) = ln6xy

Answer 1

We could do some implicit differentiation here, with chain rule applied to natural log

Answer 2

thanks Sam, i somehow found the answer in my textbook. Its really long, i'll save you the typing time

Answer 3

\[\frac{d}{dx}(5y)=5\frac{dy}{dx} \\ \\ \frac{d}{dx}(8x^4)=32x^3 \\ \\ \frac{d}{dx}(\ln(6xy))=\frac{1}{6xy}\frac{d}{dx}(6xy)=\frac{1}{6xy}(6(x\frac{dy}{dx})+y(1)) =\frac{1}{y}\frac{dy}{dx}+\frac{1}{x}\]
So, from here, substitute it in
\[5y + 8x^{4} = \ln(6xy) \\ \\ 5\frac{dy}{dx}+32x^3=\frac{1}{y}\frac{dy}{dx}+\frac{1}{x} \\ \\ \frac{dy}{dx}(5-\frac{1}{y})=\frac{1}{x}-32x^3 \\ \\ \Large \frac{dy}{dx}=\frac{\frac{1}{x}-32x^3}{5-\frac{1}{y}} \]
Then finish up from here, can you?

Answer 4

Are you a math major?

Answer 5

nope, undergrad :)

Answer 6

um but you can say that too :)

Answer 7

Lol, you're on the right track. I'm an undergrad too. Math can be fun, its mostly difficult.

Answer 8

Yeah just practice alot and it'll be alright